Quotient of locally free sheaf is locally free?

Question

If $0\rightarrow F\rightarrow G\rightarrow H \rightarrow 0$ is an extension of $\mathcal{O}$-modules with $F$ and $G$ locally free (each of constant finite rank, i.e. vector bundles), then is $H$ locally free? The same question can be asked with the roles of $F$, $G$, and $H$ interchanged, too.

In other words, is a quotient of a locally free sheaf by a subsheaf locally free?

Answer 1

No. A locally free sheaf of finite rank on a noetherian affine scheme is a finitely-generated projective module, but $$0 \longrightarrow 2 \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \longrightarrow 0$$ is an exact sequence of $\mathbb{Z}$-modules and $\mathbb{Z} / 2 \mathbb{Z}$ is not projective.

Answer 2

The complete answer can be found in Vakil's FOAG exercise 14.2Q and 14.2R 2022 version, the result is as follows:

Let the following sequence of Quasicoherent sheaves be exact :

$$0\to \mathcal{F}' \to \mathcal{F}\to \mathcal{F}''\to 0$$

(1) if $\mathcal{F}'$ and $\mathcal{F}''$ are locally free, then so it's $\mathcal{F}$

(2) if $\mathcal{F}$ and $\mathcal{F}''$ are locally free of finite rank , then so it's $\mathcal{F}'$

(3) $\mathcal{F}'$ and $\mathcal{F}$ are locally free, $\mathcal{F}''$ needs not to be locally free.

We see the kernel of surjective morphism between locally free finite rank sheaves is locally free, but it's not true for the cokernel.

Answer 3

Zhen Lin's answer is more than adequate, but I'll add another example. Consider the following short exact sequence of sheaves on $\Bbb{P}^1$: $$ 0\to \mathcal{O}(-1)\xrightarrow{\times f} \mathcal{O}\to k(x)\to 0. $$ Here, $f \in \Gamma(\Bbb{P}^1,\mathcal{O}(1)).$ This is the sequence realizing $\mathcal{O}(-1)$ as the ideal sheaf of a point. The skyscraper sheaf $k(x)$ is a torsion sheaf and hence not locally free.

This example indicates the geometric intuition for why a quotient of locally free sheaves is not locally free. Indeed, here is an even more basic example (basically the above example in local coordinates): consider the trivial line bundle over $\Bbb{C}$, just viewed as a product: $\pi:\Bbb{C}^2\to \Bbb{C}$ by $(x,y)\mapsto x$. Consider the bundle map given by $(x,y)\mapsto (x,xy)$. Fibre by fibre, the multiplication map $y\mapsto xy$ is an isomorphism except in the fibre $\pi^{-1}(0)$, where there is a nontrivial kernel and cokernel. This is the sense in which the quotient of a map of vector bundles could well be a skyscraper sheaf.